Question

If \[f\left( x \right) = \left\{
  \dfrac{{1 - \cos 4x}}{{{x^2}}},when\ x < 0 \\
  a,when\ x = 0 \\
  \dfrac{{\sqrt x }}{{\sqrt {\left( {16 + \sqrt x } \right)} - 4}},when\ x > 0 \\
  \right.\]
is continuous at \[x = 0\], then the value of \[a\] will be
A. \[8\]
B. \[ - 8\]
C. \[4\]
D. None of these

Answer 1

Hint:Here to find the value of \[a\], as it is continuous at \[x = 0\] let us apply the definition of continuity.
\[\mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right) = f\left( a \right)\]
As the function is continuous at \[x = 0\] apply this equation.
In which we need to find \[\mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right)\] and \[\mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right)\] to get the value of \[a\].

Formula used:
\[\mathop {\lim }\limits_{x \to 0} f\left( x \right) = \mathop {\lim }\limits_{x \to 0} f\left( x \right) = f\left( a \right)\]

Complete step-by-step answer:
The given function \[f\left( x \right)\] is
\[f\left( x \right) = \left\{
  \dfrac{{1 - \cos 4x}}{{{x^2}}},when\ x < 0 \\
  a,when\ x = 0 \\
  \dfrac{{\sqrt x }}{{\sqrt {\left( {16 + \sqrt x } \right)} - 4}},when\ x > 0 \\
  \right.\]
 Which is continuous at \[x = 0\].
Applying the continuity function for \[x < 0\] and \[x > 0\] for function \[f\left( x \right)\] we get
\[\mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right) = f\left( a \right)\]


In which, when \[x < 0\], the function is
 \[\dfrac{{1 - \cos 4x}}{{{x^2}}}\]
We get
\[\mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {0^ - }} \dfrac{{1 - \cos 4x}}{{{x^2}}}\]
\[ = \mathop {\lim }\limits_{x \to {0^ - }} \left( {\dfrac{{2{{\sin }^2}2x}}{{{{\left( {2x} \right)}^2}}}} \right) \cdot 4\]
Hence after simplifying the terms we get
\[\mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right) = 8\]
Now let us find the terms of
\[\mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {0^ + }} \dfrac{{\sqrt x }}{{\sqrt {\left( {16 + \sqrt x } \right)} - 4}}\]
After simplifying the terms of the function, we get
\[ = \mathop {\lim }\limits_{x \to {0^ + }} \sqrt {\left( {16 + \sqrt x } \right)} + 4\]
Hence,
\[\mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right) = 8\]
Also,
\[f\left( 0 \right) = 8\]
The value of \[a\] is \[8\].
So, the correct answer is “Option A”.

Additional Information:
The continuity of a real function \[\left( f \right)\] on a subset of the real numbers is defined when the function exists at point c and is given as
\[\mathop {\lim }\limits_{x \to c} f\left( x \right) = f\left( c \right)\]
A real function \[\left( f \right)\]is said to be continuous if it is continuous at every point in the domain of \[f\].
Consider a function \[f\left( x \right)\], and the function is said to be continuous at every point in \[\left[ {a,b} \right]\]including the endpoints a and b.
Continuity of “\[f\]” at a means,
\[\mathop {\lim }\limits_{x \to a} f\left( x \right) = f\left( a \right)\]
Continuity of “\[f\]” at b means,
\[\mathop {\lim }\limits_{x \to b} f\left( x \right) = f\left( b \right)\]

Note: To find the value of any given term, we need to see that the given function is continuous at \[x = 0\] or \[y = 0\] with respect to the given function and the conditions of \[x\]. Hence based on this we need to find the values of the given function.